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Imagine you have two shift-invariant measures $\mu, \nu$ in the Bernoulli space $\{0,1\}^{\mathbb{N}}$ with positive entropy and both are not the Bernoulli measure $(\frac{1}{2},\frac{1}{2})$. I know that:

$$ H( \mu \ast \nu) \geq \max(H( \mu ),H( \nu)) $$

Question. How can I prove this result with $>$ instead of $\geq$?

Thanks for your attention.

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  • $\begingroup$ Only you to know, the group structure I'm considering here is the Z_{2}^{\N}, given two sequences x and y, the sequence x+y is the sequence: x+y = (x1+y1 mod2, x2+y2 mod2,...) $\endgroup$ – Bruno Brogni Uggioni Jun 10 '15 at 17:04
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It is easy on some particular cases. Hope these simple cases could be generalized.

Suppose that $p\in[0,1]$ and call by $\mu^{(p)}$ the $(p,1-p)$-Bernoulli measure. Let $0\leq q,p\leq 1.$ Then $\mu^{(p)}*\mu^{(q)}=\mu^{(pq+(1-p)(1-q))}.$ In particular, if $q=0.5$ (or $p=0.5$) then $\mu^{(p)}*\mu^{(q)}=\mu^{(0.5)},$ and if $q=0,$ then $\mu^{(p)}*\mu^{(q)}=\mu^{(p)}$ (this is always true, but nice to check in this example). Using this observation, suppose that $0<q<p<1$ and $q,p\neq 0.5,$ then (an easy calculation gives that) $$H(\mu^{(p)}*\mu^{(q)})>\max\{H(\mu^{(p)}),H(\mu^{(q)})\}.$$

It is fun to consider $\epsilon>0$ small, and consider the function $$[0,1]\ni p\mapsto F(p):=H(\mu^{(p)}*\mu^{(\epsilon)}).$$ It has a unique maximum at $p=0.5$ with $F(0.5)=\log 2.$ It is monotone increasing on $[0,0.5],$ monotone decreasing on $[0.5,1]$ and $F(0)=F(1)=H(\mu^{(\epsilon)}).$

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  • $\begingroup$ Dear @user39115, you are completely right, once that when we do the convolution of two Bernoulli measures we still have a Bernoulli measure. But, for instance, this does not work with Markov measures. I don't know, for instance, if the convolution of two Markov measures will have finite memory... but thanks for your answer $\endgroup$ – Bruno Brogni Uggioni Nov 12 '15 at 16:06

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